Which one of the following general electronic configurations correctly represent a transition metal element?

Q: What is the answer to CDS (II) 2025 GK Q37?

D. Correct — D, (n – 1) d¹⁻¹⁰ ns⁰⁻². Two things have to be right in a transition-metal configuration, and this option is the only one that gets both. First, the d orbitals being filled belong to the shell one below the outermost — the (n – 1) shell, not (n – 2), and not the same shell as the outer s electrons. Scandium, the first transition metal, is 3d¹ 4s², so the d electrons sit in shell 3 while the s electrons sit in shell 4. Second, the outer s count has to allow a zero. NCERT gives the general configuration as (n – 1)d¹⁻¹⁰ ns¹⁻², then names the exception in the same sentence: palladium is 4d¹⁰ 5s⁰, with nothing in the outer s orbital at all. Writing the range as ns⁰⁻² takes that exception in its stride, which is why this option is the correct general statement rather than merely the usual one. The reason such exceptions exist at all is worth carrying away: the (n – 1)d and ns orbitals are extremely close in energy, so electrons move between them for small gains in stability — chromium is 3d⁵ 4s¹ and copper 3d¹⁰ 4s¹ for the same reason. That same near-degeneracy is what gives the transition metals their variable oxidation states, their coloured ions and their catalytic behaviour. (Provisional — pending the official UPSC key.)

(a)(n – 2) d¹⁻¹⁰ ns²
(b)(n – 2) f¹⁻¹⁴ (n – 1) d⁰⁻¹ ns²
(c)ns² np⁶ nd¹⁻¹⁰
(d)(n – 1) d¹⁻¹⁰ ns⁰⁻²

Answer

Why

Correct — D, (n – 1) d¹⁻¹⁰ ns⁰⁻². Two things have to be right in a transition-metal configuration, and this option is the only one that gets both. First, the d orbitals being filled belong to the shell one below the outermost — the (n – 1) shell, not (n – 2), and not the same shell as the outer s electrons. Scandium, the first transition metal, is 3d¹ 4s², so the d electrons sit in shell 3 while the s electrons sit in shell 4. Second, the outer s count has to allow a zero. NCERT gives the general configuration as (n – 1)d¹⁻¹⁰ ns¹⁻², then names the exception in the same sentence: palladium is 4d¹⁰ 5s⁰, with nothing in the outer s orbital at all. Writing the range as ns⁰⁻² takes that exception in its stride, which is why this option is the correct general statement rather than merely the usual one. The reason such exceptions exist at all is worth carrying away: the (n – 1)d and ns orbitals are extremely close in energy, so electrons move between them for small gains in stability — chromium is 3d⁵ 4s¹ and copper 3d¹⁰ 4s¹ for the same reason. That same near-degeneracy is what gives the transition metals their variable oxidation states, their coloured ions and their catalytic behaviour.

Why the others are wrong

(a)(n – 2) d¹⁻¹⁰ ns² — Puts the d electrons two shells below the outermost. No transition metal is built that way — scandium is 3d¹ 4s², with only one shell between the d and the s. The (n – 2) label belongs to the f orbitals, not the d.
(b)(n – 2) f¹⁻¹⁴ (n – 1) d⁰⁻¹ ns² — This is NCERT's formula for the inner transition elements, the f-block: the lanthanoids and actinoids, in which (n – 2)f orbitals are progressively filled. It describes a real and important family — just not the one the stem asks about.
(c)ns² np⁶ nd¹⁻¹⁰ — Places the d electrons in the same shell as the outer s and p electrons. That cannot happen in the ground state of a transition metal: the ns orbital is filled before the (n – 1)d orbitals begin to fill, and there is no nd occupancy alongside a completed np subshell in these elements.

Concept

The d-block occupies groups 3 to 12 of the periodic table, where d orbitals are progressively filled across each of four long periods — the 3d series from scandium to zinc, the 4d series, the 5d series and the 6d series. Because the d orbitals being filled lie one shell in from the outermost, their general outer configuration is written (n – 1)d¹⁻¹⁰ ns¹⁻², with palladium's 4d¹⁰ 5s⁰ as the standard exception. The f-block, filled in the (n – 2)f orbitals, holds the inner transition elements, the lanthanoids and actinoids. Because the (n – 1)d and ns levels are so close in energy, electrons from both take part in bonding, which is the source of the variable oxidation states, paramagnetism, coloured ions, catalytic activity and complex formation that distinguish these metals.

The fastest way through the option list is to check the shell label before anything else. Two options put the d electrons in the wrong place at once — (a) in the (n – 2) shell and (c) in the outermost shell — so they can go without further thought. That leaves the f-block formula and the d-block formula, and telling those apart is a matter of noticing whether an f subshell appears. One point of strictness is worth flagging honestly, because a careful student will spot it: IUPAC defines a transition element as one whose atom has an incompletely filled d subshell, or which can give a cation with one, a definition that excludes zinc, cadmium and mercury, whose configuration is (n – 1)d¹⁰ ns². The printed range d¹⁻¹⁰ therefore describes the d-block as a whole rather than the transition elements in that stricter sense. It is NCERT's own wording, and it is what the option is being marked against.

Key facts

NCERT gives the general outer configuration of the transition elements as (n – 1)d¹⁻¹⁰ ns¹⁻², with palladium the exception at 4d¹⁰ 5s⁰.
The (n – 1) label means the d orbitals being filled lie one shell inside the outermost s orbital — scandium is 3d¹ 4s².
The inner transition or f-block elements follow (n – 2)f¹⁻¹⁴ (n – 1)d⁰⁻¹ ns², which is exactly what option (b) states.
Chromium is 3d⁵ 4s¹ and copper 3d¹⁰ 4s¹ because half-filled and completely filled d sets are extra stable and the two levels are close in energy.
Zinc, cadmium and mercury have (n – 1)d¹⁰ ns² with completely filled d orbitals in the ground state and in their common oxidation states, and on the strict definition are not counted as transition elements.

Which shell holds the electrons being filled

s-block→The outermost ns orbital is being filled — groups 1 and 2. Sodium is 3s¹, calcium 4s².

p-block→The outermost np orbital is being filled, after ns is complete — groups 13 to 18.

d-block (transition)→The (n – 1)d orbitals, one shell inside the outermost, are filled: (n – 1)d¹⁻¹⁰ ns⁰⁻². Scandium 3d¹ 4s², chromium 3d⁵ 4s¹, palladium 4d¹⁰ 5s⁰.

f-block (inner transition)→The (n – 2)f orbitals, two shells inside the outermost: (n – 2)f¹⁻¹⁴ (n – 1)d⁰⁻¹ ns². The lanthanoids and actinoids. This is option (b).

Why the exceptions exist→(n – 1)d and ns lie very close in energy, so an electron will shift between them to reach a half-filled or completely filled d set — and the same closeness gives these metals their variable oxidation states.

Fix the shell label first: (n – 1) for d, (n – 2) for f. Two options fail on that test alone.

Study next

Writing ground-state configurations with the Aufbau principle, Hund's rule and the Pauli exclusion principleWhy chromium and copper depart from the expected filling orderGeneral properties of transition metals — variable oxidation states, coloured ions, catalysis, alloy and complex formationLanthanoids and actinoids, lanthanoid contraction, and why the f-block is set below the tableWhy zinc, cadmium and mercury are usually excluded from the transition elements

Common traps

Writing the d electrons in the outermost shell; they belong to the shell one below it.
Mistaking the f-block formula for the d-block one because both begin with a bracketed shell label.
Insisting the outer s count must be 1 or 2; palladium's empty 5s is exactly why a range starting at zero is the better general statement.

Either as a which-configuration-is-correct item like this one, or by giving an atomic number and asking for the configuration, the group or the block.

Related PYQs

NDA_GAT_2022_I_Q842022Open & attempt →

For an element with atomic number 35, which one of the following will be the correct number of electrons in its valence shell based on Bohr's model of an atom?

(a) 1
(b) 3
(c) 5
(d) 7

Answer(d) 7

The same skill applied to a single element rather than to a block. Building the configuration for atomic number 35 and reading off the outermost shell is the operation this question asks you to perform in general terms.

Practice

practice — not a real PYQ
The general outer electronic configuration of the inner transition (f-block) elements is
- (a)(n – 1)d¹⁻¹⁰ ns¹⁻²
- (b)(n – 2)f¹⁻¹⁴ (n – 1)d⁰⁻¹ ns²
- (c)ns¹⁻²
- (d)ns² np¹⁻⁶
Answer(b) (n – 2)f¹⁻¹⁴ (n – 1)d⁰⁻¹ ns² — the lanthanoids and actinoids fill f orbitals two shells inside the outermost, which is what makes them inner transition elements.
practice — not a real PYQ
Which one of the following transition elements has a completely filled d subshell and an empty outermost s orbital in its ground state?
- (a)Chromium
- (b)Copper
- (c)Palladium
- (d)Scandium
Answer(c) Palladium — its configuration is 4d¹⁰ 5s⁰, the standard exception NCERT names. Chromium is 3d⁵ 4s¹ and copper 3d¹⁰ 4s¹, both with one outer s electron.